Question

Calculate [H3O^-], [ClO4^-], and [OH^-] in an aqueous solution that is 0.165 M in HClO4(aq) at 25 degree C.

[H^+]= ???? M

[ClO4^-]= ??? M

[OH^-]= ???? M

Is the solution acidic, basic or neutral?

Answer 1

HClO4 -----------------> H+ + ClO4- (H+ = H3O+) Since HClO4 dissociates 100 percent, the concentration of H+ and ClO- will both be 0.165 mol/L

So you have the concentration of [H3O+] = [ClO4-] = 0.165 mol/L

The pH of this is -log(0.165) = 0.78

pH + pOH = 14

So the pOH = 14 - 0.78 = 13.22

You should realize that the concentration of OH^-1 can be calculated from Kw.
Kw = [H3O+][OH-]
At 25 deg C the value of Kw = 1x10^-14
Therefore [HO-] = 1x10^-14 / 0.165 = 6.06x10^-14